Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 15 - Chemical Equilibrium - Exercises - Page 665: 15.56a


$[PH_3]$= 0.0432 mol/L $[BCl_3]$= 0.0432 mol/L

Work Step by Step

Kc = $\frac{products}{reactants}$ = $\frac{[PH_3][BCl_3]}{1}$ We denote the reactants as (1) because the reactant $PH_3$$BCl_3$$(s)$ is a solid. Solids and liquids are not included in equilibrium constants. Step 1: $Kc = [PH_3][BCl_3]$ Therefore: $1.87\times 10^{-3} = [PH_3][BCl_3]$ Step 2: In the reaction the mole ratio for the products $[PH_3][BCl_3]$ is one to one. Therefore: We can denote $[PH_3][BCl_3]$ as $[x][x]$. Step 3: $[x][x]$ = $x^{2}$ So: $1.87\times 10^{-3} = x^{2}$ Using algebraic laws for exponents, we see that by squaring both sides of the equation the answer is: $x = 0.0432$ $mol/L$ Rounded to three significant digits.
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